Question: $y=x^2\tan(2x^5)$ Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $14x^8\sec^2(2x^7)$ (Choice B) B $2x^7\sec^2(x)+14x^6\tan(x)$ (Choice C) C $2x\tan(2x^5)+x^2\sec^2(2x^5)$ (Choice D) D $2x\tan(2x^5)+10x^6\sec^2(2x^5)$
Explanation: $x^2\tan(2x^5)$ is a product of a composite function and another function. Let... $u(x)=x^2$ $v(x)=\tan(x)$ $w(x)=2x^5$... then $y=u(x)\cdot v\Bigl(w(x)\Bigr)$. To find $\dfrac{dy}{dx}$, we will need to use the product rule and the chain rule! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u(x)\cdot v\Bigl(w(x)\Bigr)\right] \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot\dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \gray{\text{Product rule}} \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=2x$ $v'(x)=\sec^2(x)$ $w'(x)=10x^4$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}u'(x)\cdot{ v\Bigl(w(x)\Bigr)}+u(x)\cdot{ v'\Bigl(w(x)\Bigr)}\cdot w'(x) \\\\ &=2x\cdot{\tan(2x^5)}+x^2\cdot{\sec^2(2x^5)}\cdot 10x^4 \\\\ &=2x\tan(2x^5)+10x^6\sec^2(2x^5) \end{aligned}$ In conclusion: $\dfrac{dy}{dx}=2x\tan(2x^5)+10x^6\sec^2(2x^5)$